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Let $(A_n)$ be any sequence events and $P$ a Probability, is it true that:

$$ \lim_{n \to \infty} P(A_n) = P(\lim_{n \to \infty} A_n). $$

Many books focus on monotone sequences of events, but say nothing regarding pairwise disjoint sequences or any sequence in general. I'm interested in the latter two cases here, but decided to ask about the stronger one, since it covers the pairwise disjoint case.

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  • $\begingroup$ $\lim A_n$ does not always exist. $\endgroup$ Commented Oct 19, 2020 at 12:29

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If $\lim A_n$ exists the this is true. By Fatou's Lemma $P(\lim \inf A_n) \leq \lim \inf P(A_n)$. Applying this to $(A_n^{c})$ we see that $P(\lim \sup A_n) \geq \lim \sup P(A_n)$. If $\lim A_n$ exists the $\lim \sup A_n=\lim \inf A_n$ so we get $$\lim \sup P(A_n)$$ $$ \leq P(\lim \sup A_n)$$ $$=P(\lim \inf A_n) $$ $$ \leq \lim \inf P(A_n) $$ $$\leq \lim \sup P(A_n)$$ which proves the result.

Alternative proof: If $\lim A_n$ exists then $I_{A_n} \to I_{\lim A_n}$ and we can use DCT to get $\int \lim I_{A_n}dP =\lim \int I_{A_n}dP=\lim P(A_n)$.

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